HCF and LCM by Prime Factorisation Questions in English

(C) To find the time when they meet again at the starting point,we need to find the Least Common Multiple $(LCM)$ of the time taken by Kushan and Rajan to complete one round.
Time taken by Kushan = $18$ minutes.
Time taken by Rajan = $12$ minutes.
Prime factorization of $18 = 2 \times 3^2$.
Prime factorization of $12 = 2^2 \times 3$.
$LCM(18, 12) = 2^2 \times 3^2 = 4 \times 9 = 36$.
Therefore,they will meet again at the starting point after $36$ minutes.

(N/A) To determine if the rational number $\frac{15}{1600}$ has a terminating decimal expansion,we first simplify the fraction.
$\frac{15}{1600} = \frac{3}{320}$.
The denominator is $320 = 32 \times 10 = 2^5 \times 2 \times 5 = 2^6 \times 5^1$.
Since the denominator is of the form $2^n \times 5^m$,where $n=6$ and $m=1$ are non-negative integers,the rational number has a terminating decimal expansion.
To find the decimal expansion,we write $\frac{3}{2^6 \times 5^1} = \frac{3 \times 5^5}{2^6 \times 5^6} = \frac{3 \times 3125}{10^6} = \frac{9375}{1000000} = 0.009375$.

(A) To determine if the rational number $\frac{13}{3125}$ has a terminating decimal expansion,we examine the prime factorization of the denominator.
Step $1$: Find the prime factorization of $3125$.
$3125 = 5 \times 5 \times 5 \times 5 \times 5 = 5^5$.
Step $2$: $A$ rational number $\frac{p}{q}$ has a terminating decimal expansion if the prime factorization of $q$ is of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers.
Here,$q = 5^5 = 2^0 \times 5^5$. Since the denominator is in the form $2^n \times 5^m$,the rational number has a terminating decimal expansion.
Step $3$: To find the decimal expansion,multiply the numerator and denominator by $2^5$ to make the denominator a power of $10$.
$\frac{13}{3125} = \frac{13 \times 2^5}{5^5 \times 2^5} = \frac{13 \times 32}{10^5} = \frac{416}{100000} = 0.00416$.

(0.7) rational number $\frac{p}{q}$ has a terminating decimal expansion if the prime factorization of the denominator $q$ is of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers.
Here,the denominator is $50 = 2^1 \times 5^2$.
Since the prime factorization is in the form $2^n \times 5^m$,the rational number $\frac{35}{50}$ has a terminating decimal expansion.
To find the decimal expansion: $\frac{35}{50} = \frac{35 \times 2}{50 \times 2} = \frac{70}{100} = 0.7$.

(D) To find the $\text{g.c.d.}$ of $196$ and $38220$ using Euclid's division algorithm,we express the larger number as $a = bq + r$,where $0 \le r < b$.
Step $1$: Divide $38220$ by $196$.
$38220 = 196 \times 195 + 0$.
Since the remainder is $0$,the divisor at this stage is the $\text{g.c.d.}$.
Therefore,the $\text{g.c.d.}$ of $196$ and $38220$ is $196$.

(A) To find the $\text{g.c.d.}$ of $867$ and $255$ using Euclid's division algorithm,we follow these steps:
Step $1$: Since $867 > 255$,we apply Euclid's division lemma to $867$ and $255$:
$867 = 255 \times 3 + 102$
Step $2$: Since the remainder $102 \neq 0$,we apply the lemma to $255$ and $102$:
$255 = 102 \times 2 + 51$
Step $3$: Since the remainder $51 \neq 0$,we apply the lemma to $102$ and $51$:
$102 = 51 \times 2 + 0$
Since the remainder is now $0$,the divisor at this stage is the $\text{g.c.d.}$.
Therefore,the $\text{g.c.d.}$ of $867$ and $255$ is $51$.

(N/A) Step $1$: Find the prime factorization of the given numbers.
$510 = 2 \times 3 \times 5 \times 17$
$92 = 2^2 \times 23$
Step $2$: To find the $\text{g.c.d.}$,take the product of the smallest power of each common prime factor.
$\text{g.c.d.} (510, 92) = 2^1 = 2$
Step $3$: To find the $\text{l.c.m.}$,take the product of the highest power of each prime factor involved.
$\text{l.c.m.} (510, 92) = 2^2 \times 3^1 \times 5^1 \times 17^1 \times 23^1 = 4 \times 3 \times 5 \times 17 \times 23 = 23460$

(N/A) Step $1$: Find the prime factorization of $72$ and $90$.
$72 = 2^3 \times 3^2$
$90 = 2^1 \times 3^2 \times 5^1$
Step $2$: Calculate the $\text{g.c.d.}$ (Greatest Common Divisor) by taking the product of the smallest power of each common prime factor.
$\text{g.c.d.}(72, 90) = 2^1 \times 3^2 = 2 \times 9 = 18$
Step $3$: Calculate the $\text{l.c.m.}$ (Least Common Multiple) by taking the product of the highest power of each prime factor involved.
$\text{l.c.m.}(72, 90) = 2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5 = 360$

(N/A) To find the $\text{g.c.d.}$ and $\text{l.c.m.}$ using the fundamental theorem of arithmetic,we first find the prime factorization of each number:
$12 = 2^2 \times 3^1$
$15 = 3^1 \times 5^1$
$21 = 3^1 \times 7^1$
The $\text{g.c.d.}$ is the product of the smallest power of each common prime factor:
$\text{g.c.d.}(12, 15, 21) = 3^1 = 3$
The $\text{l.c.m.}$ is the product of the highest power of each prime factor involved:
$\text{l.c.m.}(12, 15, 21) = 2^2 \times 3^1 \times 5^1 \times 7^1 = 4 \times 3 \times 5 \times 7 = 420$
Thus,the $\text{g.c.d.}$ is $3$ and the $\text{l.c.m.}$ is $420$.

(A) To find the $\text{g.c.d.}$ and $\text{l.c.m.}$ using the fundamental theorem of arithmetic,we first find the prime factorization of each number:
$8 = 2^3$
$9 = 3^2$
$25 = 5^2$
$\text{g.c.d.}$ is the product of the smallest power of each common prime factor. Since there are no common prime factors,the $\text{g.c.d.}$ is $1$.
$\text{l.c.m.}$ is the product of the highest power of each prime factor involved:
$\text{l.c.m.} = 2^3 \times 3^2 \times 5^2$
$\text{l.c.m.} = 8 \times 9 \times 25$
$\text{l.c.m.} = 72 \times 25 = 1800$.

(N/A) rational number $\frac{p}{q}$ has a terminating decimal expansion if the prime factorization of the denominator $q$ is of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers.
Here,the denominator is $200$.
The prime factorization of $200$ is $200 = 2^3 \times 5^2$.
Since the denominator is in the form $2^n \times 5^m$,the rational number $\frac{23}{200}$ has a terminating decimal expansion.
To find the decimal expansion,we can write the denominator as a power of $10$:
$\frac{23}{200} = \frac{23 \times 5}{200 \times 5} = \frac{115}{1000} = 0.115$.

(A) To find the least number of chocolates that are equal for both brands,we need to calculate the Least Common Multiple $(LCM)$ of $24$ and $15$.
Prime factorization of $24 = 2^3 \times 3^1$.
Prime factorization of $15 = 3^1 \times 5^1$.
$LCM(24, 15) = 2^3 \times 3^1 \times 5^1 = 8 \times 3 \times 5 = 120$.
To find the number of boxes for the first brand ($24$ chocolates per pack): $120 / 24 = 5$ boxes.
To find the number of boxes for the second brand ($15$ chocolates per pack): $120 / 15 = 8$ boxes.
Therefore,one needs to buy $5$ boxes of the first kind and $8$ boxes of the second kind.

(A) To find the largest number that divides $615$ and $963$ leaving a remainder of $6$ in each case,we first subtract the remainder from both numbers.
$615 - 6 = 609$
$963 - 6 = 957$
Now,we need to find the Highest Common Factor $(HCF)$ of $609$ and $957$.
Prime factorization of $609 = 3 \times 7 \times 29$
Prime factorization of $957 = 3 \times 11 \times 29$
The common factors are $3$ and $29$.
$HCF = 3 \times 29 = 87$.
Therefore,the largest number is $87$.

(B) To find the largest number that divides $626$,$3127$,and $15628$ leaving remainders $1$,$2$,and $3$ respectively,we subtract the remainders from the respective numbers:
$626 - 1 = 625$
$3127 - 2 = 3125$
$15628 - 3 = 15625$
Now,we need to find the Highest Common Factor $(HCF)$ of $625$,$3125$,and $15625$.
Prime factorization of $625 = 5^4$
Prime factorization of $3125 = 5^5$
Prime factorization of $15625 = 5^6$
The $HCF$ is the product of the lowest power of common prime factors,which is $5^4 = 625$.
Therefore,the largest number is $625$.

(C) To find the longest rod that can measure the three dimensions ($15\,m$,$12\,m$,and $21\,m$) exactly,we need to calculate the Highest Common Factor $(HCF)$ of these three numbers.
Step $1$: Find the prime factorization of each dimension:
$15 = 3 \times 5$
$12 = 2^2 \times 3$
$21 = 3 \times 7$
Step $2$: Identify the common prime factors.
The only common prime factor among $15$,$12$,and $21$ is $3$.
Step $3$: The $HCF$ is the product of the lowest powers of common prime factors.
$HCF(15, 12, 21) = 3$.
Therefore,the longest rod that can measure the dimensions exactly is $3\,m$.

(D) To find the smallest five-digit number divisible by $15$,$55$,and $99$,we first find the Least Common Multiple $(LCM)$ of these numbers.
Step $1$: Prime factorization of the numbers:
$15 = 3 \times 5$
$55 = 5 \times 11$
$99 = 3^2 \times 11$
Step $2$: Calculate the $LCM$ by taking the highest power of each prime factor present:
$LCM(15, 55, 99) = 3^2 \times 5 \times 11 = 9 \times 5 \times 11 = 495$.
Step $3$: The smallest five-digit number is $10000$.
Divide $10000$ by $495$:
$10000 \div 495 = 20.202...$
Step $4$: To find the smallest five-digit multiple,we take the next integer quotient,which is $21$.
$495 \times 21 = 10395$.
Thus,the smallest five-digit number divisible by $15$,$55$,and $99$ is $10395$.

(A) To find when the bells toll together,we need to calculate the Least Common Multiple $(LCM)$ of $8$,$12$,$15$,and $18$.
Prime factorization:
$8 = 2^3$
$12 = 2^2 \times 3$
$15 = 3 \times 5$
$18 = 2 \times 3^2$
$LCM = 2^3 \times 3^2 \times 5 = 8 \times 9 \times 5 = 360$ seconds.
This means the bells toll together every $360$ seconds,which is $360 / 60 = 6$ minutes.
In one hour ($60$ minutes),the number of times they toll together is $60 / 6 = 10$ times.
Since we exclude the toll at the start,the total count remains $10$.

(B) Let two consecutive natural numbers be $n$ and $n+1$,where $n \in \mathbb{N}$.
Any common divisor of $n$ and $n+1$ must also divide their difference.
The difference is $(n+1) - n = 1$.
The only positive divisor of $1$ is $1$.
Therefore,the greatest common divisor $(\text{g.c.d.})$ of $n$ and $n+1$ is $1$.

(D) We know that for any two positive integers $a$ and $b$,the relationship between their $\text{g.c.d.}$ (Greatest Common Divisor) and $\text{l.c.m.}$ (Least Common Multiple) is given by:
$\text{g.c.d.}(a, b) \times \text{l.c.m.}(a, b) = a \times b$
Given that $\text{g.c.d.} = 18$ and the product of the two numbers $(a \times b) = 6480$.
Substituting these values into the formula:
$18 \times \text{l.c.m.} = 6480$
$\text{l.c.m.} = \frac{6480}{18}$
$\text{l.c.m.} = 360$
Therefore,the $\text{l.c.m.}$ of the two numbers is $360$.

(A) By definition,a prime number is a natural number greater than $1$ that has no positive divisors other than $1$ and itself.
Since $k_{1}$ and $k_{2}$ are two distinct prime integers,they share no common factors other than $1$.
For any two numbers $a$ and $b$,the relationship between their $\text{H.C.F.}$ and $\text{L.C.M.}$ is given by the product of the numbers: $a \times b = \text{H.C.F.}(a, b) \times \text{L.C.M.}(a, b)$.
Since $k_{1}$ and $k_{2}$ are distinct primes,their $\text{H.C.F.}(k_{1}, k_{2}) = 1$.
Therefore,$k_{1} \times k_{2} = 1 \times \text{L.C.M.}(k_{1}, k_{2})$.
Thus,$\text{L.C.M.}(k_{1}, k_{2}) = k_{1} k_{2}$.

(C) To find the $\text{g.c.d.}$ (Greatest Common Divisor) of $24$ and $63$,we perform prime factorization of both numbers:
$24 = 2^3 \times 3^1$
$63 = 3^2 \times 7^1$
The $\text{g.c.d.}$ is the product of the smallest power of each common prime factor.
The common prime factor is $3$.
The smallest power of $3$ present in both is $3^1 = 3$.
Therefore,$\text{g.c.d.}$ $(24, 63) = 3$.

(D) To find the $\text{l.c.m.}$ of $36$ and $94$,we first find their prime factorizations:
$36 = 2^2 \times 3^2$
$94 = 2 \times 47$
The $\text{l.c.m.}$ is the product of the highest power of each prime factor present in the numbers:
$\text{l.c.m.} = 2^2 \times 3^2 \times 47$
$\text{l.c.m.} = 4 \times 9 \times 47$
$\text{l.c.m.} = 36 \times 47 = 1692$

(D) The greatest common divisor (g.c.d.) of $12$ and $40$ is calculated as follows:
Prime factorization of $12 = 2^2 \times 3^1$.
Prime factorization of $40 = 2^3 \times 5^1$.
The g.c.d. is the product of the smallest powers of common prime factors: $2^2 = 4$.
Given the equation: $\text{g.c.d.}(12, 40) = 40 + 4x$.
Substituting the value: $4 = 40 + 4x$.
Subtracting $40$ from both sides: $4 - 40 = 4x$.
$-36 = 4x$.
Dividing by $4$: $x = -9$.

(A) To find the decimal form of the fraction $\frac{6}{15}$,we first simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor,which is $3$.
$\frac{6 \div 3}{15 \div 3} = \frac{2}{5}$.
Now,to convert $\frac{2}{5}$ into a decimal,we can multiply both the numerator and the denominator by $2$ to make the denominator $10$:
$\frac{2 \times 2}{5 \times 2} = \frac{4}{10} = 0.4$.
Therefore,the decimal form of $\frac{6}{15}$ is $0.4$.

(A) We know that for any two positive integers $a$ and $b$,the relationship between their greatest common divisor $(GCD)$ and least common multiple $(LCM)$ is given by the formula:
$\text{GCD}(a, b) \times \text{LCM}(a, b) = a \times b$
Given $a = 336$ and $b = 52$.
First,we find the prime factorization of $336$ and $52$:
$336 = 2^4 \times 3 \times 7$
$52 = 2^2 \times 13$
The $\text{GCD}(336, 52) = 2^2 = 4$.
Now,using the formula:
$4 \times \text{LCM}(336, 52) = 336 \times 52$
$\text{LCM}(336, 52) = \frac{336 \times 52}{4}$
$\text{LCM}(336, 52) = 84 \times 52 = 4368$
Thus,the correct option is $A$.

(D) To find the $L.C.M.$ of $115$ and $25$,we first find their prime factorizations:
$115 = 5 \times 23$
$25 = 5 \times 5 = 5^2$
The $L.C.M.$ is the product of the highest power of each prime factor present in the numbers:
$L.C.M.(115, 25) = 5^2 \times 23$
$L.C.M.(115, 25) = 25 \times 23 = 575$

(C) To find the Greatest Common Divisor ($G$.$C$.$D$.) of $28$,$35$,and $91$,we first find the prime factorization of each number:
$28 = 2^2 \times 7$
$35 = 5 \times 7$
$91 = 7 \times 13$
The $G$.$C$.$D$. is the product of the smallest power of each common prime factor present in all the numbers.
Here,the only common prime factor is $7$,and its smallest power is $7^1$.
Therefore,$G$.$C$.$D$. $(28, 35, 91) = 7$.

(B) For any two positive integers $a$ and $b$,the fundamental theorem of arithmetic establishes a relationship between their $G$.$C$.$D$. and $L$.$C$.$M$.
Specifically,the product of the $G$.$C$.$D$. $(a, b)$ and the $L$.$C$.$M$. $(a, b)$ is equal to the product of the two numbers themselves.
Therefore,$\text{G.C.D.}(a, b) \times \text{L.C.M.}(a, b) = a \times b = ab$.

(A) To find the $LCM$ of $15, 21,$ and $35$,we first find the prime factorization of each number:
$15 = 3 \times 5$
$21 = 3 \times 7$
$35 = 5 \times 7$
The $LCM$ is the product of the highest power of each prime factor present in the numbers:
$LCM(15, 21, 35) = 3^1 \times 5^1 \times 7^1 = 3 \times 5 \times 7 = 105$
Therefore,the correct option is $A$.

(C) To find the $LCM$ of $40, 60,$ and $80$,we first find the prime factorization of each number:
$40 = 2^3 \times 5^1$
$60 = 2^2 \times 3^1 \times 5^1$
$80 = 2^4 \times 5^1$
The $LCM$ is the product of the highest power of each prime factor present in the numbers:
$LCM = 2^4 \times 3^1 \times 5^1$
$LCM = 16 \times 3 \times 5$
$LCM = 16 \times 15 = 240$

(D) To find the smallest number that leaves a remainder of $5$ when divided by $20, 30,$ and $40,$ we first find the Least Common Multiple $(LCM)$ of the divisors.
$LCM(20, 30, 40)$:
$20 = 2^2 \times 5$
$30 = 2 \times 3 \times 5$
$40 = 2^3 \times 5$
$LCM = 2^3 \times 3 \times 5 = 8 \times 3 \times 5 = 120$.
The number must be of the form $(LCM \times k) + \text{remainder}$.
Here,$LCM = 120$ and $\text{remainder} = 5$.
So,the number $= 120k + 5$.
For the smallest positive number greater than $5$,we take $k = 1$.
Number $= 120(1) + 5 = 125$.

(C) To find the smallest positive number divisible by $24, 36$,and $48$,we need to calculate the Least Common Multiple $(LCM)$ of these numbers.
Step $1$: Find the prime factorization of each number:
$24 = 2^3 \times 3^1$
$36 = 2^2 \times 3^2$
$48 = 2^4 \times 3^1$
Step $2$: The $LCM$ is the product of the highest power of each prime factor present in the numbers:
$LCM = 2^4 \times 3^2$
$LCM = 16 \times 9 = 144$
Therefore,the required smallest number is $144$.

(D) The integers from $2$ to $6$ are $2, 3, 4, 5,$ and $6$.
To find the smallest positive number divisible by all these integers,we need to calculate the Least Common Multiple $(LCM)$ of $2, 3, 4, 5,$ and $6$.
Prime factorization of each number:
$2 = 2^1$
$3 = 3^1$
$4 = 2^2$
$5 = 5^1$
$6 = 2^1 \times 3^1$
Taking the highest power of each prime factor present:
$LCM = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60$.
Therefore,$60$ is the smallest positive number divisible by $2, 3, 4, 5,$ and $6$.

(B) To find the smallest positive number divisible by every integer from $2$ to $10$,we need to calculate the Least Common Multiple $(LCM)$ of the numbers $2, 3, 4, 5, 6, 7, 8, 9, 10$.
First,write the prime factorization of each number:
$2 = 2^1$
$3 = 3^1$
$4 = 2^2$
$5 = 5^1$
$6 = 2 \times 3$
$7 = 7^1$
$8 = 2^3$
$9 = 3^2$
$10 = 2 \times 5$
The $LCM$ is found by taking the highest power of each prime factor present in the numbers:
$LCM = 2^3 \times 3^2 \times 5^1 \times 7^1$
$LCM = 8 \times 9 \times 5 \times 7$
$LCM = 72 \times 35 = 2520$.
Therefore,the smallest positive number divisible by all integers from $2$ to $10$ is $2520$.

(C) According to the fundamental property of numbers,for any two positive integers $a$ and $b$,the product of their $\text{g.c.d.}$ and $\text{l.c.m.}$ is equal to the product of the numbers themselves.
Formula: $\text{g.c.d.}(a, b) \times \text{l.c.m.}(a, b) = a \times b$.
Here,$a = 18$ and $b = 24$.
Therefore,$\text{g.c.d.}(18, 24) \times \text{l.c.m.}(18, 24) = 18 \times 24$.
Calculating the product: $18 \times 24 = 432$.

(B) We know the fundamental relationship between the greatest common divisor (g.c.d.) and the least common multiple (l.c.m.) of two positive integers $a$ and $b$ is given by the formula:
$\text{l.c.m.} (a, b) \times \text{g.c.d.} (a, b) = a \times b$
Given that $\text{g.c.d.} (a, b) = b$,we can substitute this into the formula:
$\text{l.c.m.} (a, b) \times b = a \times b$
Dividing both sides by $b$ (since $b \in N$,$b \neq 0$):
$\text{l.c.m.} (a, b) = \frac{a \times b}{b} = a$
Therefore,the correct option is $B$.

(C) The greatest common divisor (g.c.d.) of two integers $a$ and $b$,denoted as $d = \text{gcd}(a, b)$,is the largest positive integer that divides both $a$ and $b$.
$1$. By definition,$d$ must divide both $a$ and $b$,so $d \mid a$ and $d \mid b$ is true.
$2$. Any common divisor $c$ of $a$ and $b$ must be less than or equal to the greatest common divisor $d$. Thus,$c \leqslant d$ is true.
$3$. By the property of the greatest common divisor,any common divisor $c$ of $a$ and $b$ must also divide their greatest common divisor $d$. Thus,$c \mid d$ is true.
$4$. The statement $c > d$ is false because if $c$ is a common divisor,it cannot be strictly greater than the greatest common divisor $d$.

(B) To solve this,we calculate the Least Common Multiple ($L$.$C$.$M$.) and Greatest Common Divisor ($G$.$C$.$D$.) for each pair:
$1$. $\text{l.c.m.}(8, 16, 24)$:
Prime factors: $8 = 2^3$,$16 = 2^4$,$24 = 2^3 \times 3^1$.
$L$.$C$.$M$. = $2^4 \times 3^1 = 16 \times 3 = 48$. (Matches $b$)
$2$. $\text{g.c.d.}(8, 16, 24)$:
Common factors: $8 = 2^3$,$16 = 2^4$,$24 = 2^3 \times 3^1$.
$G$.$C$.$D$. = $2^3 = 8$. (Matches $c$)
$3$. $\text{l.c.m.}(6, 12, 18)$:
Prime factors: $6 = 2 \times 3$,$12 = 2^2 \times 3$,$18 = 2 \times 3^2$.
$L$.$C$.$M$. = $2^2 \times 3^2 = 4 \times 9 = 36$. (Matches $a$)
$4$. $\text{g.c.d.}(6, 12, 18)$:
Common factors: $6 = 2 \times 3$,$12 = 2^2 \times 3$,$18 = 2 \times 3^2$.
$G$.$C$.$D$. = $2^1 \times 3^1 = 6$. (Matches $e$)
Thus,the correct matching is $(1-b), (2-c), (3-a), (4-e)$.

(D) The smallest prime number is $2$.
The smallest composite number is $4$.
To find the $LCM$ of $2$ and $4$:
$2 = 2^1$
$4 = 2^2$
$LCM(2, 4) = 2^2 = 4$.
Therefore,the $LCM$ of the smallest prime number and the smallest composite number is $4$.

(B) The $HCF$ of two numbers always divides their $LCM$ completely.
Given $HCF(a, b) = 16$.
We check each option to see if it is divisible by $16$:
$A) 32 / 16 = 2$ (Divisible)
$B) 72 / 16 = 4.5$ (Not divisible)
$C) 64 / 16 = 4$ (Divisible)
$D) 96 / 16 = 6$ (Divisible)
Since $72$ is not divisible by $16$,it cannot be the $LCM$ of $a$ and $b$.

(D) The smallest prime number is $2$.
The smallest composite number is $4$.
To find the $LCM$ of $2$ and $4$:
$2 = 2^1$
$4 = 2^2$
$LCM(2, 4) = 2^2 = 4$.
Therefore,the $LCM$ of the smallest prime number and the smallest composite number is $4$.

(C) For distinct prime numbers $a, b, c$,the Highest Common Factor $(HCF)$ is the greatest common divisor,which is $1$ because they share no common factors other than $1$.
The Least Common Multiple $(LCM)$ of distinct prime numbers is the product of the numbers themselves,which is $a \times b \times c = abc$.
Therefore,the ratio of $HCF$ to $LCM$ is $\frac{\text{HCF}}{\text{LCM}} = \frac{1}{abc}$.
This can be expressed as the ratio $1: abc$.

(A) The fundamental property of $HCF$ and $LCM$ states that the $LCM$ of two numbers must be perfectly divisible by their $HCF$.
Given $HCF(a, b) = 12$.
We check each option to see if it is divisible by $12$:
$(A)$ $90 \div 12 = 7.5$ (Not divisible)
$(B)$ $24 \div 12 = 2$ (Divisible)
$(C)$ $48 \div 12 = 4$ (Divisible)
$(D)$ $36 \div 12 = 3$ (Divisible)
Since $90$ is not divisible by $12$, it cannot be the $LCM(a, b)$.

(B) First,find the prime factorization of $65$ and $117$.
$65 = 5 \times 13$
$117 = 9 \times 13 = 3^2 \times 13$
The $\text{HCF}(65, 117)$ is the common factor with the lowest power,which is $13$.
Given the equation: $\text{HCF}(65, 117) = 65m - 117$
Substitute the value of $\text{HCF}$:
$13 = 65m - 117$
Add $117$ to both sides:
$13 + 117 = 65m$
$130 = 65m$
Divide by $65$:
$m = \frac{130}{65} = 2$
Thus,the value of $m$ is $2$.

(C) The $LCM$ (Least Common Multiple) of two numbers is the smallest positive integer that is divisible by both numbers.
Since $p$ and $q$ are distinct prime numbers,they have no common factors other than $1$.
Therefore,the $LCM$ of two distinct prime numbers is simply their product.
Thus,$LCM(p, q) = p \times q = pq$.

(A) To find the $LCM(26, 91)$,we first find the prime factorization of each number:
$26 = 2 \times 13$
$91 = 7 \times 13$
The $LCM$ is the product of the highest power of each prime factor present in the numbers:
$LCM(26, 91) = 2 \times 7 \times 13 = 182$
Therefore,the correct option is $A$.

(D) prime number is a natural number greater than $1$ that has no positive divisors other than $1$ and itself.
Since $m$ and $n$ are distinct prime integers,they do not share any common factors other than $1$.
The Highest Common Factor $(HCF)$ of two distinct prime numbers is always $1$ because their only common divisor is $1$.
Therefore,the $HCF$ of $m$ and $n$ is $1$.

(B) For any two positive integers $a$ and $b$,the $LCM(a, b)$ must be perfectly divisible by their $HCF(a, b)$.
Given that $HCF(a, b) = 25$,the $LCM(a, b)$ must be a multiple of $25$.
Checking the given options:
$(A)$ $50 = 25 \times 2$ (Possible)
$(B)$ $105$ is not divisible by $25$ $(105 / 25 = 4.2)$ (Impossible)
$(C)$ $100 = 25 \times 4$ (Possible)
$(D)$ $25 = 25 \times 1$ (Possible)
Therefore,$105$ cannot be the $LCM$ of $a$ and $b$.

(A) We know that for any two positive integers $a$ and $b$,the relationship $a \times b = LCM(a, b) \times GCD(a, b)$ holds true.
Given that $LCM(a, b) = a \times b$.
Substituting this into the formula,we get $a \times b = (a \times b) \times GCD(a, b)$.
Dividing both sides by $a \times b$ (assuming $a, b \neq 0$),we get $GCD(a, b) = 1$.
Therefore,the correct answer is $1$.

(A) To find the $HCF$ (Highest Common Factor) of $12$,$15$,and $21$,we first find the prime factorization of each number:
$12 = 2^2 \times 3^1$
$15 = 3^1 \times 5^1$
$21 = 3^1 \times 7^1$
The $HCF$ is the product of the smallest power of each common prime factor present in the numbers.
The only common prime factor is $3$,and its smallest power is $3^1$.
Therefore,$HCF$ $(12, 15, 21) = 3$.